Some Applications of the Chain Rule

The Chain Rule will help us determine the derivatives of logarithms and exponential functions $\mathrm{a}^{x}$ . We will also use it to answer some applied questions and to find slopes of graphs given by parametric equations.

Source: Dale Hoffman, https://s3.amazonaws.com/saylordotorg-resources/wwwresources/site/wp-content/uploads/2012/12/MA005-3.6-Some-Applications-of-the-Chain-Rule.pdf
This work is licensed under a Creative Commons Attribution 3.0 License.

$\mathbf{D}(\ln (x))=\dfrac{1}{x} \quad$ and $\quad \mathbf{D}(\ln (g(x)))=\dfrac{\mathbf{g}^{\prime}(x)}{g(x)}$

Proof: We know that the natural logarithm $\ln (x)$ is the logarithm with base $\mathrm{e}$, and $e^{\ln (x)}=x$ for $x > 0$

We also know that $\mathbf{D}\left(\mathrm{e}^{\mathrm{X}}\right)=\mathrm{e}^{\mathrm{x}}$, so using the Chain Rule we have $\mathrm{D}\left(\mathrm{e}^{\mathrm{f}(\mathrm{x})}\right)=\mathrm{e}^{\mathrm{f}(\mathrm{x})} \mathbf{f}^{\prime}(\mathbf{x}) .$ Differentiating each side of the equation $\mathrm{e}^{\ln (\mathrm{x})}=\mathrm{x}$, we get that

The function $\ln (\mathrm{g}(\mathrm{x}))$ is the composition of $\mathrm{f}(\mathrm{x})=\ln (\mathrm{x})$ with $\mathrm{g}(\mathrm{x})$, so by the Chain Rule,

$D \big((\ln (g(x))=D(f(g(x)))=f^{\prime}(g(x)) \cdot g^{\prime}(x)=\frac{1}{g(x)} \cdot g^{\prime}(x)=\frac{g^{\prime}(x)}{g(x)}\big).$

Example 1: Find $\mathbf{D}(\ln (\sin (x)))$ and $\mathbf{D}\left(\ln \left(x^{2}+3\right)\right)$.

Solution: (a) Using the pattern $\mathbf{D}\big((\ln (\mathrm{g}(\mathrm{x}))=\frac{\boldsymbol{g}^{\prime}(\mathrm{x})}{\mathrm{g}(\mathrm{x})} \big)$ with $\mathrm{g}(\mathrm{x})=\sin (\mathrm{x})$, then

$\mathbf{D}(\ln (\sin (x)))=\frac{g^{\prime}(x)}{g(x)}=\frac{D(\sin (x))}{\sin (x)}=\frac{\cos (x)}{\sin (x)}=\cot (x)$

(b) Using the pattern with $\mathrm{g}(\mathrm{x})=\mathrm{x}^{2}+3$, we have $\mathrm{D}\left(\ln \left(\mathrm{x}^{2}+3\right)\right)=\frac{\mathrm{g}^{\prime}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}=\frac{2 \mathrm{x}}{\mathrm{x}^{2}+3}$.

We can use the Change of Base Formula from algebra to rewrite any logarithm as a natural logarithm, and then we can differentiate the resulting natural logarithm.

Change of Base Formula for logarithms: $\log _{\mathrm{a}} \mathrm{x}=\frac{\log _{\mathrm{b}} \mathrm{x}}{\log _{\mathrm{b}} \mathrm{a}} \quad$ for all positive $\mathrm{a}, \mathrm{b}$ and $\mathrm{x}$.

Example 2: Use the Change of Base formula and your calculator to find $\log _{\pi} 7$ and $\log _{2} 8$.

Solution: $\log _{\pi} 7=\frac{\ln 7}{\ln \pi} \approx \frac{1.946}{1.145} \approx 1.700 .$ (Check that $\left.\pi^{1.7} \approx 7\right) \log _{2} 8=\frac{\ln 8}{\ln 2}=3$.

Practice 1: Find the values of $\log _{9} 20, \log _{3} 20$ and $\log _{\pi}$ $\mathrm{e}$.

Putting $b=e$ in the Change of Base Formula, $\log _{a} x=\frac{\log _{e} x}{\log _{e} a}=\frac{\ln x}{\ln a}$, so any logarithm can be written as a natural logarithm divided by a constant. Then any logarithm is easy to differentiate.

$\mathbf{D}\left(\log _{a}(x)\right)=\frac{1}{x \ln (a)} \quad$ and $D\left(\log _{a}(f(x))\right)=\frac{f^{\prime}(x)}{f(x)} \cdot \frac{1}{\ln (a)}$

The second differentiation formula follows from the Chain Rule.

Practice 2: Calculate $\mathbf{D}\left(\log _{10}(\sin (x))\right)$ and $\mathbf{D}\left(\log_{\pi}\left(\mathrm{e}^{\mathrm{x}}\right)\right)$.

The number $\mathrm{e}$ might seem like an "unnatural" base for a natural logarithm, but of all the logarithms to different bases, the logarithm with base e has the nicest and easiest derivative. The natural logarithm is even related to the distribution of prime numbers. In 1896, the mathematicians Hadamard and Valle-Poussin proved the following conjecture of Gauss: (The Prime Number Theorem) For large values of $\mathrm{x}, \{$ number of primes less than $\mathrm{x}\} \approx \frac{\mathrm{X}}{\ln (\mathrm{x})}$.

 Once we know the derivative of $\mathrm{e}^{\mathrm{X}}$ and the Chain Rule, it is relatively easy to determine the derivative of $a^{x}$ for any $a > 0$.

$D\left(a^{x}\right)=a^{x} \cdot \ln a \quad$ for $a > 0$.

Proof: If $a > 0$, then $a^{x} > 0$ and $a^{x}=e^{\ln \left(a^{x}\right)}=e^{x \cdot \ln a}$.

$D\left(a^{x}\right)=D\left(e^{\ln \left(a^{x}\right)}\right)=D\left(e^{x \cdot \ln a}\right)=e^{x \cdot \ln a} \cdot D(x \cdot \ln a)=a^{x} \cdot \ln \mathbf{a}$.

Example 3: Calculate $\mathbf{D}\left(7^{\mathrm{X}}\right)$ and $\frac{\mathbf{d}}{\mathbf{d t}}\left(2^{\sin (\mathrm{t})}\right)$

Solution: (a) $\mathrm{D}\left(7^{\mathrm{X}}\right)=7^{\mathrm{x}} \ln 7 \approx(1.95) 7^{\mathrm{x}}$.

(b) We can write $\mathrm{y}=2^{\sin (\mathrm{t})}$ as $\mathrm{y}=2^{\mathrm{u}}$ with $\mathrm{u}=\sin (\mathrm{t})$. Using the Chain Rule,

$\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dt}}=2^{\mathrm{u}} \cdot \ln (2) \cdot \cos (\mathrm{t})=2^{\sin (\mathrm{t})} \cdot \ln (2) \cdot \cos (\mathrm{t})$.

Practice 3: $\quad$ Calculate $\mathbf{D}\left(\sin \left(2^{\mathrm{X}}\right)\right)$ and $\frac{\mathrm{d}}{\mathrm{dt}}\left(3^{\left(\mathrm{t}^{2}\right)}\right)$.

Now we can examine applications which involve more complicated functions.

Example 4: A ball at the end of a rubber band (Fig. 1) is oscillating up and down, and its height 

(in feet) above the floor at time $t$ seconds is $h(t)=5+2 \sin (t / 2)$. (t is in radians)

(a) How fast is the ball travelling after 2 seconds? after 4 seconds? after 60 seconds?

(b) Is the ball moving up or down after 2 seconds? after 4 seconds? after 60 seconds?

(c) Is the vertical velocity of the ball ever $0$?

Solution: (a) $v(t)=D(h(t))=D(5+2 \sin (t / 2))$

                $=2 \cos (t / 2) \mathbf{D}(t / 2)=\cos (t / 2)$ feet/second so

                $v(2)=\cos (2 / 2) \approx 0.540 \mathrm{ft} / \mathrm{s}, \mathrm{v}(4)=\cos (4 / 2) \approx-0.416 \mathrm{ft} / \mathrm{s}$, and

                $v(60)=\cos (60 / 2) \approx 0.154 \mathrm{ft} / \mathrm{s}$

(b) The ball is moving upward when $\mathrm{t}=2$ and 60 seconds, downward when $\mathrm{t}=4$.

(c) $v(t)=\cos (t / 2)$ and $\cos (t / 2)=0$ when $t=\pi \pm n \cdot 2 \pi \quad(n=1,2, \ldots)$.

Example 5: If 2400 people now have a disease, and the number of people with the disease appears to double every 3 years, then the number of people expected to have the disease in $\mathrm{t}$ years is $\mathrm{y}=2400 \cdot 2^{\mathrm{t} / 3}$.

(a) How many people are expected to have the disease in 2 years?

(b) When are 50,000 people expected to have the disease?

(c) How fast is the number of people with the disease expected to grow now and 2 years from now?

Solution: (a) In 2 years, $y=2400 \cdot 2^{2 / 3} \approx 3,810$ people.

(b) We know $\mathrm{y}=50,000$, and we need to solve $50,000=2400 \cdot 2^{\mathrm{t} / 3}$ for $\mathrm{t}$. Taking logarithms of each side of the equation, $\ln (50,000)=\ln \left(2400 \cdot 2^{t / 3}\right)=\ln (2400)+(t / 3) \cdot \ln (2)$ so $10.819=7.783+.231 \mathrm{t} \quad$ and $\mathrm{t} \approx$ 13.14 years. We expect 50,000 people to have the disease about 13.14 years from now.

(c) This is asking for $\mathrm{dy} / \mathrm{dt}$ when $\mathrm{t}=0$ and 2 years. $\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{d}\left(2400 \cdot 2^{\mathrm{t} / 3}\right)}{\mathrm{dt}}=2400 \cdot 2^{\mathrm{t} / 3} \cdot \ln (2) \cdot(1 / 3)$ $\approx 554.5 \cdot 2^{\mathrm{t} / 3}$. Now, at $\mathrm{t}=0$, the rate of growth of the disease is approximately $554.5 \cdot 2^{0} \approx 554.5$ people/year. In 2 years the rate of growth will be approximately $554.5 \cdot 2^{2 / 3} \approx 880$ people/year.

Example 6: You are riding in a balloon, and at time $\mathrm{t}$ (in minutes) you are $\mathrm{h}(\mathrm{t})=\mathrm{t}+\sin (\mathrm{t})$ feet high. If the temperature at an elevation $\mathrm{h}$ is $\mathrm{T}(\mathrm{h})=\frac{72}{1+\mathrm{h}} \quad$ degrees Fahrenheit, then how fast is your temperature changing when $\mathrm{t}=5$ minutes? (Fig. 2)

Solution: As $\mathrm{t}$ changes, your elevation will change, and, as your elevation changes, so will your temperature. It is not difficult to write the temperature as a function of time, and then we could calculate

$\frac{\mathrm{d} \mathrm{T}(\mathrm{t})}{\mathrm{dt}}=\mathrm{T}^{\prime}(\mathrm{t})$ and evaluate $\mathrm{T}^{\prime}(5)$, or we could use the Chain Rule:

$\frac{\mathbf{d} \mathrm{T}(\mathrm{t})}{\mathbf{d} \mathbf{t}}=\frac{\mathbf{d} \mathrm{T}(\mathrm{h}(\mathrm{t}))}{\mathbf{d} \mathrm{h}(\mathrm{t})} \cdot \frac{\mathrm{d} \mathrm{h}(\mathrm{t})}{\mathrm{dt}}=\frac{\mathbf{d} \mathrm{T}(\mathrm{h})}{\mathbf{d} \mathrm{h}} \cdot \frac{\mathrm{d} \mathrm{h}(\mathrm{t})}{\mathrm{dt}}=\frac{-72}{(1+\mathrm{h})^{2}} \cdot(1+\cos (\mathrm{t}))$

When $t=5$, then $h(t)=5+\sin (5) \approx 4.04$ so $T^{\prime}(5) \approx \frac{-72}{(1+4.04)^{2}} \cdot(1+.284) \approx-3.64^{\circ} /$ minute.

Practice 4: Write the temperature $\mathrm{T}$ in the previous example as a function of the variable $\mathrm{t}$ alone and then differentiate $\mathrm{T}$ to determine the value of $\mathrm{dT} / \mathrm{dt}$ when $\mathrm{t}=5$ minutes.

Example 7: A scientist has determined that, under optimum conditions, an initial population of 40 bacteria will grow "exponentially" to $\mathrm{f}(\mathrm{t})=40 \cdot \mathrm{e}^{\mathrm{t} / 5}$ bacteria after $\mathrm{t}$ hours.

(a) Graph $\mathrm{y}=\mathrm{f}(\mathrm{t})$ for $0 \leq \mathrm{t} \leq 15$. Calculate $\mathrm{f}(0), \mathrm{f}(5), \mathrm{f}(10)$.

(b) How fast is the population increasing at time $t$ ? (Find $f^{\prime}(t)$).

(c) Show that the rate of population increase, $f^{\prime}(t)$, is proportional to the population, $f(t)$, at any time t. $\quad$ (Show $\mathrm{f}^{\prime}(\mathrm{t})=\mathrm{K} \cdot \mathrm{f}(\mathrm{t})$ for some constant $\mathrm{K}$).

Solution: (a) The graph of $\mathrm{y}=\mathrm{f}(\mathrm{t})$ is given in Fig. 3. $\mathrm{f}(0)=40 \cdot \mathrm{e}^{0 / 5}=40$ bacteria. $\mathrm{f}(5)=40 \cdot \mathrm{e}^{5 / 5} \approx 109$ bacteria and $f(10)=40 \cdot \mathrm{e}^{10 / 5} \approx 296$ bacteria.

(b) $f^{\prime}(t)=\frac{\mathbf{d}}{d t}(f(t))=\frac{\mathbf{d}}{d t}\left(40 \cdot e^{\mathrm{t} / 5}\right)=40 \cdot \mathrm{e}^{\mathrm{t} / 5} \frac{\mathbf{d}}{\mathbf{d} t}(\mathrm{t} / 5)$

$=40 \cdot \mathrm{e}^{\mathrm{t} / 5}(1 / 5)=8 \cdot \mathrm{e}^{\mathrm{t} / 5}$ bacteria/hour.

(c) $\quad \mathrm{f}^{\prime}(\mathrm{t})=8 \cdot \mathrm{e}^{\mathrm{t} / 5}=\frac{1}{5} \cdot\left(40 \cdot \mathrm{e}^{\mathrm{t} / 5}\right)=\frac{1}{5} \mathrm{f}(\mathrm{t})$ so $\mathrm{f}^{\prime}(\mathrm{t})=\mathrm{K} \cdot \mathrm{f}(\mathrm{t})$ with $\mathrm{K}=1 / 5$.

Suppose a robot has been programmed to move in the $\mathrm{xy}$-plane so at time $\mathrm{t}$ its $\mathrm{x}$ coordinate will be $\sin (t)$ and its $\mathrm{y}$ coordinate will be $t^{2}$. Both $x$ and $y$ are functions of the independent parameter t, $x(t)=\sin (t)$ and $y(t)=t^{2}$, and the path of the robot (Fig. 4) can be found by plotting $(x, y)=(x(t), y(t))$ for lots of values of $t$.

$\begin{array}{c|c|c|c} \mathrm{t} & \mathrm{x}(\mathrm{t})=\sin (\mathrm{t}) & \mathrm{y} \mathrm{t})=\mathrm{t}^{2} & \text { plot point at} \\ \hline 0 & 0 & 0 & (0,0) \\ .5 & .48 & .25 & (.48, .25) \\ 1.0 & .84 & 1 & (.84,1) \\ 1.5 & 1.00 & 2.25 & (1,2.25) \\ 2.0 & .91 & 4 & (.91,4) \end{array}$ 

Typically we know $\mathrm{x}(\mathrm{t})$ and $\mathrm{y}(\mathrm{t})$ and need to find $\mathrm{dy} / \mathrm{dx}$, the slope of the tangent line to the graph of $(\mathrm{x}(\mathrm{t}), \mathrm{y}(\mathrm{t}))$. The Chain Rule says that $\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{dy}}{\mathrm{dx}} \cdot \frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}$, so, algebraically solving for $\frac{\mathrm{dy}}{\mathrm{dx}}$, we get $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy} / \mathrm{dt}}{\mathrm{dx} / \mathrm{dt}}$.

If we can calculate $\mathrm{dy} / \mathrm{dt}$ and $\mathrm{dx} / \mathrm{dt}$, the derivatives of $\mathrm{y}$ and $\mathrm{x}$ with respect to the parameter $\mathrm{t}$, then we can determine $\mathrm{dy/dx}$, the rate of change of $\mathrm{y}$ with respect to $\mathrm{x}$.

Example 8: Find the slope of the tangent line to the graph of $(x, y)=\left(\sin (t), t^{2}\right)$ when $t=2 ?$
Solution: $\mathrm{dx} / \mathrm{dt}=\cos (\mathrm{t})$ and $\mathrm{dy} / \mathrm{dt}=2 \mathrm{t} .$ When $\mathrm{t}=2$, the object is at the point $\left(\sin (2), \mathbf{2}^{2}\right) \approx(.91,4)$ and the slope of the tangent line to the graph is $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy} / \mathrm{dt}}{\mathrm{dx} / \mathrm{dt}}=\frac{2 \mathrm{t}}{\cos (\mathrm{t})}=\frac{2 \cdot 2}{\cos (\mathbf{2})} \approx \frac{4}{-.42} \approx-9.61$.

Practice 5: Graph $(x, y)=(3 \cos (t), 2 \sin (t))$ and find the slope of the tangent line when $t=\pi/2$.

When we calculated $\frac{d y}{d x}$, the slope of the tangent line to the graph of $(\mathrm{x}(\mathrm{t}), \mathrm{y}(\mathrm{t}))$, we used the derivatives $\frac{d x}{d t}$ and $\frac{d y}{d t}$, and each of these derivatives also has a geometric meaning:

$\frac{d x}{d t}$ measures the rate of change of $\mathrm{x}(\mathrm{t})$ with respect to $\mathrm{t}$ - it tells us whether the $\mathrm{x}$-coordinate is increasing or decreasing as the t-variable increases.

$\frac{d y}{d t}$ measures the rate of change of $\mathrm{y}(\mathrm{t})$ with respect to $\mathrm{t}$.

Example 9: For the parametric graph in Fig. 5, tell whether $\frac{d x}{d t}, \frac{d y}{d t}$ and $\frac{d y}{d x}$ is positive or negative when $\mathrm{t=2}$.

Solution: As we move through the point $\mathrm{B}$ (where $\mathrm{t}=2$ ) in the direction of increasing values of $t$, we are moving to the left so $x(t)$ is decreasing and $\frac{d x}{d t}$ is negative.

Similarly, the values of $\mathrm{y}(\mathrm{t})$ are increasing so $\frac{d y}{d t}$ is positive. Finally, the slope of the tangent line, $\frac{d y}{d x}$, is negative.

(As check on the sign of $\frac{d y}{d x}$ we can also use the result $\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{\text { positive }}{\text { negative}}$  = $\text{negative}$.)

Practice 6: For the parametric graph in the previous example, tell whether $\frac{d x}{d t}, \frac{d y}{d t}$ and $\frac{d y}{d x}$ is positive or negative when $\mathrm{t}=1$ and when $\mathrm{t}=3$.

If we know the position of an object at every time, then we can determine its speed. The formula for speed comes from the distance formula and looks a lot like it, but with derivatives.

Proof: The speed of an object is the limit, as $\Delta \mathrm{t} \rightarrow 0$, of $\frac{\text { change in position }}{\text { change in time }}$. (Fig. 6 )

$\begin{gathered} \frac{\text { change in position }}{\text { change in time }}=\frac{\sqrt{(\Delta x)^{2}+(\Delta y)^{2}}}{\Delta t}=\sqrt{\frac{(\Delta x)^{2}+(\Delta y)^{2}}{(\Delta t)^{2}}} \\ \qquad \qquad \qquad \qquad =\sqrt{\left(\frac{\Delta x}{\Delta t}\right)^{2}+\left(\frac{\Delta y}{\Delta t}\right)^{2}} \\ \quad \qquad \qquad \qquad \qquad \qquad \qquad \rightarrow \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} \qquad \text { as } \Delta t \rightarrow 0 \end{gathered}$

Exercise 10: Find the speed of the object whose location at time $\mathrm{t}$ is $(\mathrm{x}, \mathrm{y})=\left(\sin (\mathrm{t}), \mathrm{t}^{2}\right)$ when $\mathrm{t}=0$ and $\mathrm{t}=1$.

Solution: $\mathrm{dx} / \mathrm{dt}=\cos (\mathrm{t})$ and $\mathrm{dy} / \mathrm{dt}=2 \mathrm{t}$ so speed $=\sqrt{(\cos (\mathrm{t}))^{2}+(2 \mathrm{t})^{2}}=\sqrt{\cos ^{2}(\mathrm{t})+4 \mathrm{t}^{2}}$.

Practice 7: Show that an object whose location at time $t$ is $(x, y)=(3 \sin (t), 3 \cos (t))$ has a constant speed. (This object is moving on a circular path).

Practice 8: Is the object whose location at time $\mathrm{t}$ is $(\mathrm{x}, \mathrm{y})=(3 \cos (\mathrm{t}), 2 \sin (\mathrm{t}))$ travelling faster at the top of the ellipse ( at $\mathrm{t}=\pi / 2$ ) or at the right edge of the ellipse (at $\mathrm{t}=0$)?

Practice 1: $\log _{9} 20=\frac{\log (20)}{\log (9)} \approx 1.3634165 \approx \frac{\ln (20)}{\ln (9)}, \log _{3} 20=\frac{\log (20)}{\log (3)} \approx 2.726833 \approx \frac{\ln (20)}{\ln (3)}$

                   $\log _{\pi} \mathrm{e}=\frac{\log (\mathrm{e})}{\log (\pi)} \approx \mathbf{0 . 8 7 3 5 6 8 5} \approx \frac{\ln (\mathrm{e})}{\ln (\pi)}=\frac{1}{\ln (\pi)}$

Practice 2: $\mathbf{D}\left(\log _{10}(\sin (\mathrm{x}))\right)=\frac{1}{\sin (\mathrm{x}) \cdot \ln (10)} \quad \mathbf{D}(\sin (\mathrm{x}))=\frac{\cos (\mathbf{x})}{\sin (\mathrm{x}) \cdot \ln (10)}$

                 $\mathbf{D}\left(\log _{\pi}\left(\mathrm{e}^{\mathrm{x}}\right)\right)=\frac{1}{\mathrm{e}^{\mathrm{x}} \cdot \ln (\pi)} \mathrm{D}\left(\mathrm{e}^{\mathrm{x}}\right)=\frac{\mathbf{e}^{\mathbf{x}}}{\mathrm{e}^{\mathrm{x}} \cdot \ln (\pi)}=\frac{1}{\ln (\pi)}$

Practice 3: $D\left(\sin \left(2^{x}\right)\right)=\cos \left(2^{x}\right) D\left(2^{x}\right)=\cos \left(2^{x}\right) \cdot 2^{x} \cdot \ln (2)$

                 $\frac{\mathbf{d}}{\mathbf{d t}} 3^{\left(\mathrm{t}^{2}\right)}=3^{\left(\mathrm{t}^{2}\right)} \ln (3) \mathbf{D}\left(\mathrm{t}^{2}\right)=3^{\left(\mathrm{t}^{2}\right)} \ln (3) \cdot 2 \mathrm{t}$

Practice 4: $\mathrm{T}=\frac{72}{1+\mathrm{h}}=\frac{72}{1+\mathrm{t}+\sin (\mathrm{t})}$

                  $\frac{\mathbf{d} \mathrm{T}}{\mathbf{d} \mathbf{t}}=\frac{(1+\mathrm{t}+\sin (\mathrm{t})) \cdot \mathbf{D}(72)-72 \cdot \mathbf{D}(1+\mathrm{t}+\sin (\mathrm{t}))}{(1+\mathrm{t}+\sin (\mathrm{t}))^{2}}=\frac{-72(1+\cos (\mathrm{t}))}{(1+\mathrm{t}+\sin (\mathrm{t}))^{2}}$

                  When $t=5, \frac{\mathbf{d} T}{\mathbf{d t}}=\frac{-72(1+\cos (5))}{(1+\mathrm{t}+\sin (5))^{2}} \approx-\mathbf{3 . 6 3 6 9 5}$.

Practice 5: $\mathrm{x}(\mathrm{t})=3 \cos (\mathrm{t})$ so $\mathrm{dx} / \mathrm{dt}=-3 \sin (\mathrm{t}) \cdot \mathrm{y}(\mathrm{t})=2 \sin (\mathrm{t})$

                 so $\mathrm{dy} / \mathrm{dt}=2 \cos (\mathrm{t}) \cdot \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy} / \mathrm{dt}}{\mathrm{dx} / \mathrm{dt}}=\frac{2 \cos (\mathrm{t})}{-3 \sin (\mathrm{t})}$.

                  When $t=\pi / 2, \frac{d y}{d x}=\frac{2 \cos (\pi / 2)}{-3 \sin (\pi / 2)}=\frac{2 \cdot 0}{-3 \cdot 1}=0 .$ (See Fig. 12)

Practice 6: When $x=1$ : pos., pos., pos. When $x=3$ : pos., neg., neg.

Practice 7: $\mathrm{x}(\mathrm{t})=3 \sin (\mathrm{t})$ and $\mathrm{y}(\mathrm{t})=3 \cos (\mathrm{t})$ so $\mathrm{dx} / \mathrm{dt}=3 \cos (\mathrm{t})$ and $\mathrm{dy} / \mathrm{dt}=-3 \sin (\mathrm{t}).$ Then

Practice 8: $\mathrm{x}(\mathrm{t})=3 \cos (\mathrm{t})$ and $\mathrm{y}(\mathrm{t})=2 \sin (\mathrm{t})$ so $\mathrm{dx} / \mathrm{dt}=-3 \sin (\mathrm{t})$ and $\mathrm{dy} / \mathrm{dt}=2 \cos (\mathrm{t}) .$ Then